Prove that each of the four triangles formed by joining the mid-points of the sides of a triangle are similar to the…

Let D, E, F be the mid-points of sides BC, CA, AB respectively of $\triangle ABC$.
By Mid-point Theorem, DE $||$ AB and DE $= \frac{1}{2}$ AB.
EF $||$ BC and EF $= \frac{1}{2}$ BC.
FD $||$ AC and FD $= \frac{1}{2}$ AC.
Consider $\triangle AFE$ and $\triangle ABC$.
$\frac{AF}{AB} = \frac{1}{2}$ and $\frac{AE}{AC} = \frac{1}{2}$ (F and E are mid-points)
$\angle A$ is common.
So, $\triangle AFE \sim \triangle ABC$ (SAS similarity criterion).
Similarly, $\triangle BDF \sim \triangle ABC$ and $\triangle CED \sim \triangle ABC$.
Also, DE $||$ AB, so ADEF is a parallelogram.
$\angle FDE = \angle A$ (Opposite angles of parallelogram)
$\frac{FD}{AC} = \frac{1}{2}$, $\frac{DE}{AB} = \frac{1}{2}$, $\frac{FE}{BC} = \frac{1}{2}$
So, $\triangle FDE \sim \triangle ABC$ (SSS similarity criterion).
Thus, all four triangles are similar to the original triangle.